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Materials 101 Problem Set #41.) The tensile stress-strain curve for a cylindrical specimen of brass is reproduced in Figure 1.

A. Find the following:a) The work done to fracture per unit volumeThe work done to fracture per unit volume is the area underneath the Nominal Stress vs. Nominal

Strain curve. If we had an equation to describe this curve, we could integrate the equation from zero

to the strain at fracture. Since we dont have this equation available, estimate the work done by

estimating the area beneath the curve. Work done to fracture per unit volume ~ 20.5 boxes * 10

J/m^3/box = 205 J/m^3.

b) The Young's modulus of brassThe Youngs modulus is found from the slope of the Nominal Stress vs. Nominal Strain curve in in the

linear regime where nominal strain is very small. In this case you should estimate the slope of the

curve. E ~100 MPa/.0025 = 40,000 MPa.

c)The ultimate tensile stressThe ultimate tensile stress (UTS) occurs where the Nominal Stress vs. Nominal Strain curve shows a

maximum. In this case the UTS ~530 MPa.

d) The 0.2% offset yield stressTo find the 0.2% offset yield stress, extrapolate a line with the same slope as the linear regime (which

is equal to the Youngs modulus) from the x axis of Nominal Stress vs. Nominal Strain curve upwards

until it intersects with the curve. The stress at this intersection is the 0.2% offset yield stress. Here it

is ~ 150 MPa.e)The proportional limitThe proportional limit occurs where the stress and strain stop being linear or proportional. This is

sometimes used as a measure the yield stress (but not in this class) and is ~ 95 MPa.B. Suppose a similar specimen is given a tensile prestrain of 0.2. On subsequent testing this specimen you

would find an approximate yield stress of 492 MPa ?

When the specimen is loaded again, the stress would be proportional to the strain, starting at the 0.20

strain, until it hits the normal stress-strain curve. Applying this to the first n-n plot given, a nominalyield stress based on the original cross-sectional area Ao would be 410 MPa. However if as usual the

cross-sectional area Ao1 is measured before the second test we need to find this area in order to get the

correct stress from the value we read off the original stress-strain curve. We know this stress (the true

stress) = 410 Ao/Ao1 , where Ao/Ao1 = Lo1/Lo = 1+n = 1.20. The yield stress will be thus 492 MPa.

Another way to look at this problem is as follows: The sample will not start to yield again (undergopermanent plastic deformation) until it reaches a stress equivalent to that stress it was experiencing the

last time it was yielding. In this case yielding last occurred at a prestrain of 0.2. We know that

permanent plastic deformation occurred during the prestrain, so we know that the area of the sample

changed from Ao to Ao1.

Yield stress (y) is the point where elastic deformation stops and permanent plastic deformation begins.You can calculate the yield stress after the prestrain, by converting the nominal stress of the prestrain of

0.2 (this is the strain at which the sample last yielded) to what it would be with the new area Ao1. (As I

stated earlier, the yield stress will occur at the a stress equivalent to the stress value it last yielded at.

This calculation finds that equivalent stress.) However, as stated earlier, the sample has a new area, Ao1.

So y = F/Ao*( Ao/Ao1)= n*( Ao/Ao1). Note, this is the same thing as calculating the true stress at anominal strain of 0.2.

What is the true strain and true stress at a point of a curve in Figure 1 that corresponds to 0.35 nominal strain?At the nominal strain of 0.35 we can find the true strain by using the fact that true strain = ln(L/Lo) =

ln(1+ n ) = 0.30. To find the true stress you need to multiply the nominal stress 520 MPa by Ao/A =1+ n = 1.35. This gives a true stress of 702 MPa. Why can't you compute the realandat the nominalstrain of 0.45 from this curve? This is because at nominal strain of 0.40,where the nominal stress-nominalstrain curve is a maximum, necking starts. Hence at 0.45 nominal strain, necking has already occurred.

Because the curve is obtained by monitoring the variation of nominal strain from changes in the length,

beyond 0.40 strain the data cannot be used to calculate the true strain.

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0

100

200

300

400

500

600

0 0.1 0.2 0.3 0.4 0.5 0.6

Nominal Strain

NominalStress(MPa)

Brass (25 C)

X

0

50

100

150

200

250

0.000 0.005 0.010 0.015 0.020 0.025 0.030 0.035

NominalStress(MPa)

Low strain region of the above curve

Nominal Strain

Ultimate Tensile Stress

0.2% offset yield stress

The proportional limit

0.002

2


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equivalent


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Problem 5

5.) The true stress-strain curve for the steel in question 4 is given by

t

=897 (

t

) 0.3 (.

t

) 0.01 (stress in MPa)

i) What is the true strain at the ultimate tensile tensile stress? Assume=0 forpart i).

ii) What is the ultimate tensile stress for the steel at a t =1 sec-1?

Answer:i) The ultimate tensile stress occurs at the maximum in the force F vs.

true strain curve. Since F = tA, we can find the maximum with

true strain from dF/dt= 0 = A(dt/ d) + t(dA/dt) where A =Aoexp(-t). Taking the derivatives we find thatt UTS= 0.3 the

strain hardening coeffficient.

Additionally using the following relationship where t =Km,

= m at necking since the UTS corresponds to the point at which

the deformation starts to localize, forming a neck.

ii)

From part i) (t) at the UTS(strain) = 897(0.3)0.3(1)

0.01= 625 MPa

=t. To get the ultimate tensile stress, which is the nominal stress,

we need to multiply tby A/Ao, which comes from the following

equation. t=* (Ao/A) . So we can solve for the nominal stress

()UTS = 625 *exp(-t) = 625(.714)= 463 MPa.


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Materials 101 Problem Set #4

6.) The hotel balcony designed in Problem 4 collapses, killing 200 people. It is suspectedthat the supporting rod was the cause of (or at least a contributing factor in) the failure.You examine a failed connecting rod and find the dimensions in the sketch below. There

is no evidence of any torsional strain, implying

during the failure was zero.

0.95 inches in diameter

0.50 inches in diameter i) What was the local true strain at fracture?ii) Is the appearance of the failed supporting rod consistent with its being constructed

with the steel in part b above? Give quantitative reasoning.

Problem Set #4 Question #6

Part i)Ao = A (1+En) where En is the nominal strain.Solve for nominal strain and convert to true strain according to:E(true) = ln (1+En).

En=Ao/A-1=.95/.5-1= 0.9Et=ln(1+0.9)=0.64

Part i)

This unnecked diameter, which is reduced from theoriginal 1 inch diameter, is the key. This means that the true strain atthe UTS (the last strain at which the sample was deforming uniformly) was-2ln(0.95) = 0.102. But the strain hardening exponent given in question 3(which should equal the strain at the UTS) is given as 0.3. SO it looks asif the wrong steel was used.

1

Ao/A=(1/0.5)^2 = 4 ln(Ao/A) = 1.38

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